To Prove $T $ is a self map and $T$ have no fixed points

Question

Let $K=\{x=(x(n))\in c_0:0\le x(n)\le 1$ for all $n\in \mathbb{N}\}$. Define $T:K\to c_0$ by $T(x)=(1,x(1),x(2),x(3),...).$ Prove :

(a) $T$ is a self map on $K$ and $||Tx-Ty||_\infty=||x-y||_\infty $

(b) If $0_{c_0}\in F$ and $T(F)\subseteq F.$ where $F$ is a closed convex subset of $K$ then $x_n=e_1+e_2+...+e_n\in F$, for all $n\in \mathbb{N}$

(c) $T$ does not have any fixed point in $K$

My try:

$||.||_\infty = \sup_{i \geq 1} |a_n|$

im trying to calculate $||Tx-Ty||_{\infty}=||((0,x_1-y_1,.....)||_{\infty}=\sup_{n\ge 1}|x_n-y_n||=||x-y||_\infty$ is i am correct?

for (c) if $Tx=x$ then $(1,x_1,x_2,...)=(x_1,x_2,...)$ then $x_1=1=x_2=x_3..$

but $x_n $ not in $c_0$

and i dont know how to prove that $T$ is self MAP and i dont know how to prove (b)

Answer 1

Herein I take $c_0$ to be the sequences $(x(n))$, $n \in \Bbb N$, such that

$\displaystyle \lim_{n \to \infty} x(n) = 0; \tag 1$

that is, $c_0$ is the set of sequences which converge to $0$. I also take it that "$T$ is a self map on $K$" means that the range of $T$ lies in $K$; that is,

$T:K \to K. \tag 2$

These things being said, for

(a): if $x = (x(n)) \in K$, we are given that

$Tx = (1, x(1), x(2), \ldots ); \tag 3$

that is,

$(Tx)(1) = 1, \tag 4$

$(Tx)(n) = x(n - 1), \; \; 2 \le n \in \Bbb N; \tag 5$

since

$0 \le x(n) \le 1, \; \forall n \in \Bbb N, \tag 6$

we have

$0 \le T(x(n)) \le 1, \forall n \in \Bbb N \tag 7$

as well; thus,

$T(x(n)) \in K, \tag 8$

i.e, $T:K \to K$; $T$ is a self-map on $K$; also, with

$y = y(n) \in K, \tag 9$

$\Vert Tx - Ty \Vert_\infty = \Vert (0, x(1) - y(1), x(2) - y(2), \ldots ) \Vert_\infty$ $= \sup \{0, \vert x(n) - y(n) \vert, \; n \in \Bbb N \} = \sup \{ \vert x(n) - y(n) \vert, \; n \in \Bbb N \}, \tag{10}$

where this last equality binds by virtue of the fact that

$\vert x(n) - y(n) \vert \ge 0, \; \forall n \in \Bbb N; \tag{11}$

now

$\sup \{ \vert x(n) - y(n) \vert, \; n \in \Bbb N \} = \Vert x - y \Vert_\infty; \tag{12}$

combining (10) and (12) yields

$\Vert Tx - Ty \Vert_\infty = \Vert x - y \Vert_\infty; \tag{13}$

as for

(b): with

$0_{c_0} = (0, 0, 0 \ldots) \in F \tag{13}$

and

$T(F) \subseteq F, \tag{14}$

we have

$e_1 = (1, 0, 0, \ldots) = T0_{c_0} \in F, \tag{15}$

$e_1 + e_2 = (1, 1, 0, 0, \ldots) = Te_1 = T^20_{c_0} \in F, \tag{16}$

$e_1 + e_2 + e_3 = (1, 1, 1, 0, 0, \ldots) = T(e_1 + e_2) = T^30_{c_0} \in F; \tag{17}$

if we denote by $E_k \in K$ the sequence consisting of $k$ leading $1$s and every other element $0$, that is

$E_k(j) = 1, \; 1 \le j \le k, \tag{18}$

$E_k(j) = 0, \; j > k, \tag{19}$

we may formulate an inductive hypothesis

$e_1 + e_2 + \ldots + e_k = E_k = T(e_1 + e_2 + \ldots + e_{k - 1}) = T^k 0_{c_0} \in F, \tag{20}$

of which (15)-(17) are the first three cases $k = 1, 2, 3$; applying $T$ to this equation yields

$T(e_1 + e_2 + \ldots + e_k) = TE_k = T^2(e_1 + e_2 + \ldots + e_{k - 1}) = T^{k + 1}0_{c_0} \in F, \tag{21}$

where (14) implies the rightmost assertion

$T^{k + 1} 0_{c_0} \in F; \tag{22}$

it is evident from (20) and the definitions that

$TE_k = E_{k + 1} = E_k + e_{k + 1} = e_1 + e_2 + \ldots + e_k + e_{k + 1}, \tag{23}$

and we may use this to transform (21) into

$e_1 + e_2 + \ldots + e_k + e_{k + 1} = E_{k + 1} = T(e_1 + e_2 + \ldots + e_{k - 1} + e_k) = T^{k + 1} 0_{c_0} \in F, \tag{24}$

completing the induction; thus (20) holds for every $1 \le k \in \Bbb N$; we have thus completed the demonstration of part (b); finally, we turn to

(c): a fixed point $x \in K$ of $T$ satisfies

$T(x(n)) = x(n), \tag{25}$

whence

$T^2(x(n)) = TT(x(n)) = T(x(n)) = x(n), \tag{26}$

$T^3(x(n)) = TT^2(x(n)) = T(x(n)) = x(n); \tag{27}$

and indeed if

$T^k(x(n)) = x(n), \tag{28}$

we have

$T^{k + 1}(x(n)) = TT^k(x(n)) = T(x(n)) = x(n); \tag{29}$

thus we see inductivly that (28) binds for all $1 \le k \in \Bbb N$.

Now for

$(x(n)) \in K \subset c_0, \tag{30}$

for any $\epsilon > 0$ there exists

$0 < N \in \Bbb N \tag{31}$

such that

$j > N \Longrightarrow x(j) < \epsilon; \tag{32}$

now if we choose $k$ in (20) with

$k > N, \tag{33}$

we see by means of (18)-(19) that

$\exists n > N, \; E_k(n) = 1; \tag{34}$

but this contradicts (32); thus $T$ has no fixed points in $K$.

Note: Apparently we do not need the hypotheses that $F$ is closed and convex to attain (b). End of Note.

To Prove $T $ is a self map and $T$ have no fixed points

Answer 2

Let $x=(x_n)$ be a sequence that converges to $0$ with $\max_n x_n \leq 1.$ Let $y=(y_n)=T(x)$ Then $x_n=y_{n+1}$ and $y_0=1$, so $\max_n y_n =1$ and $\lim_{n\to \infty} y_n=\lim_{n \to \infty} x_n=0$, so $T$ maps $K$ to $K$.

If $0 \in F$ and $T(F) \subseteq F$, then $T^n(0)=\sum_{k=1}^n e_k \in F$.